Power Factor Question

[perryallen]

I know this should be easy, but I found this question and I was looking for a simple way to express the formula.

*What is the Power Factor of a motor that reads 1576 watts on a wattmeter and 5.892 on a digital amp meter and is 86 percent Effecient at 240 volts?


I was doing va=E(240v)x I(5.892A) / EFF (.86)= 1644va


PF=W(1576) / va (1644) = .958 PF

Is this correct? I can't find anything in the reference works to express this?

 

[charlie b]

Something is not looking right. Efficiency is the relationship between electrical energy supplied to the motor and the mechanical work the motor performs. It is not a player in the power factor equation. You have readings of the electrical input voltage and the electrical current being supplied. Those two are all you need to calculate VA input. However, leaving the efficiency out of the equation gives a resulting power factor that is higher than 1.0. That is not physically possible. One thing you did not mention is whether this is a single phase or three phase motor. If it is a three phase, then I get a power factor of 64.4%. Do you have the book?s answer to the question?

 

[perryallen]

Something is not looking right. Efficiency is the relationship between electrical energy supplied to the motor and the mechanical work the motor performs. It is not a player in the power factor equation. You have readings of the electrical input voltage and the electrical current being supplied. Those two are all you need to calculate VA input. However, leaving the efficiency out of the equation gives a resulting power factor that is higher than 1.0. That is not physically possible. One thing you did not mention is whether this is a single phase or three phase motor. If it is a three phase, then I get a power factor of 64.4%. Do you have the book?s answer to the question?

That was my problem. I think the question is flawed. It didn't state 3 phase or single phase. I assumed single and was getting the same figures as you so I thought maybe Effeciency should be taken into account.

Here is the link. Scroll down to sample question 37. Only thing is, look at the possible answers. Lowest is .70 http://www.electricianeducation.com/digital/master_electrician_exams_2008_nec.htm

 

[kwired]

Has to be a three phase motor even though it is not mentioned otherwise you have over 1.0 power factor as Charlie mentioned.

Efficiency marked on motor is not really related to power factor that much. It is simply how much input power results in output power at the motor shaft, at rated full load and voltage. If motor is not running at full load or at rated voltage this number will vary from what is on nameplate.

 

[Besoeker]

Something is not looking right. Efficiency is the relationship between electrical energy supplied to the motor and the mechanical work the motor performs. It is not a player in the power factor equation. You have readings of the electrical input voltage and the electrical current being supplied. Those two are all you need to calculate VA input. However, leaving the efficiency out of the equation gives a resulting power factor that is higher than 1.0. That is not physically possible. One thing you did not mention is whether this is a single phase or three phase motor. If it is a three phase, then I get a power factor of 64.4%. Do you have the book?s answer to the question?

I agree. The figures suggest three phase. And a PF of 0.64 seems plausible for such a small motor.

 

[david luchini]

The only problem with the three phase model is that 0.64 power factor is not one of the choices in the online problem linked.

 

[Besoeker]

The only problem with the three phase model is that 0.64 power factor is not one of the choices in the online problem linked.

I didn't see a link for the on line problem.
Can you copy it or just give the actual question?

 

[david luchini]

I didn't see a link for the on line problem.
Can you copy it or just give the actual question?

Look at the end of post #3.

 

[LMAO]

another remote possibility is Wattmeter is reading motor output power (mechanical power); either way motor has to be 3 phase for the problem to make any sense. if power meter is reading motor output then PF is 0.75.

 

[kwired]

another remote possibility is Wattmeter is reading motor output power; either way motor has to be 3 phase for the problem to make any sense. if power meter is reading motor output then PF is 0.75.

That would take more than just a wattmeter that measures volts and amps to come up with the output power of the motor, not saying it couldn't be done, efficiency will vary with the load so you can't just assume a constant value - the value on nameplate should be efficiency at full load with nameplate voltage applied.

 

[LMAO]

That would take more than just a wattmeter that measures volts and amps to come up with the output power of the motor, not saying it couldn't be done, efficiency will vary with the load so you can't just assume a constant value - the value on nameplate should be efficiency at full load with nameplate voltage applied.

what kind of "remote possibility" don't you understand? but seriously, that's the only way you can use the given effiecieny.

 

[jumper]

Perhaps it is time to look for a different source of training materials. :happyyes:

But it was written by these guys.

International Network of Electrical Professional Testers


http://forums.mikeholt.com/archive/index.php?t-118893.html&

 

[Besoeker]

Look at the end of post #3.

Appreciated, thank you.
So 240V and 5.892A is 1414 VA single phase. The power measured is 1576W.
Can't be single phase.
For three phase the VA would be 2423VA from sqrt(3) *240*5.892. Around 2449VA
Resulting PF is 0.64.
None of the answers give that as an option.
I like to see the the basis of their answer that excludes that.

 

[Jraef]

I think you guys are all WAY over thinking this and it's getting complicated by trying to make the values real. As I see it this is just a math problem, I doubt anyone put much thought into it.

P (W) = V x A x PF x eff.

So...

You know the W on one side, and you know the the V, the A and the eff on the other side, solve for the missing PF.

 

[kwired]

I think you guys are all WAY over thinking this and it's getting complicated by trying to make the values real. As I see it this is just a math problem, I doubt anyone put much thought into it.

P (W) = V x A x PF x eff.

So...

You know the W on one side, and you know the the V, the A and the eff on the other side, solve for the missing PF.

Unless I am doing something wrong doesn't that give you a power factor greater than 1.0 with the information we have?

 

[Besoeker]

I think you guys are all WAY over thinking this and it's getting complicated by trying to make the values real. As I see it this is just a math problem, I doubt anyone put much thought into it.

P (W) = V x A x PF x eff.

So where is that power measured?

 

[perryallen]

I think you guys are all WAY over thinking this and it's getting complicated by trying to make the values real. As I see it this is just a math problem, I doubt anyone put much thought into it.

P (W) = V x A x PF x eff.

So...

You know the W on one side, and you know the the V, the A and the eff on the other side, solve for the missing PF.

Are you saying that you would solve it like this:

240v x 5.892A x .86 Eff = 2106va

1576w / 2106va= .748 power factor (That is an option but is the math sound?)

 

[kwired]

Are you saying that you would solve it like this:

240v x 5.892A x .86 Eff = 2106va

1576w / 2106va= .748 power factor (That is an option but is the math sound?)

My algebra may not be the best but I don't think that is the correct method to find the missing element in the formula.

The formula is P = V x A x PF x eff

Plug in the known values

1576w = 240v x 5.892a x PF x .86eff

this can be re written as

1576w = 1216.1 x PF

divide both sides of equation by 1216.1

1.296 = PF

that number should not be possible in real world, with motor only and no correction added.

If this is wrong it is not the math it is the wrong formula.

 

[perryallen]

My algebra may not be the best but I don't think that is the correct method to find the missing element in the formula.

The formula is P = V x A x PF x eff

Plug in the known values

1576w = 240v x 5.892a x PF x .86eff

this can be re written as

1576w = 1216.1 x PF

divide both sides of equation by 1216.1

1.296 = PF

that number should not be possible in real world, with motor only and no correction added.

If this is wrong it is not the math it is the wrong formula.

I forot to put in the 1.732 for three phase, which I believe was probably meant in the original question. That changes to 2106, not 1216.

 

[__dan]

I know this should be easy, but I found this question and I was looking for a simple way to express the formula.

*What is the Power Factor of a motor that reads 1576 watts on a wattmeter and 5.892 on a digital amp meter and is 86 percent Effecient at 240 volts?


I was doing va=E(240v)x I(5.892A) / EFF (.86)= 1644va


PF=W(1576) / va (1644) = .958 PF

Is this correct? I can't find anything in the reference works to express this?

deleted: I see Charlie answered this in post #2